IB Mathematics AA – Topic 2: Functions
Comprehensive Guide to Modulus Functions & Inequalities
Introduction to Modulus Functions
The modulus function (also called the absolute value function) measures the distance of a number from zero on the number line, regardless of direction. It always produces a non-negative output, making it essential for modeling situations involving magnitude, distance, error, and tolerance.
Key concept: The modulus of \(x\), written as \(|x|\), gives the "size" or "magnitude" of \(x\) without considering its sign. Whether \(x\) is positive or negative, \(|x|\) is always non-negative. This seemingly simple function has profound applications in optimization, physics, engineering, and computer science.
Why modulus matters: Modulus functions create V-shaped graphs with distinct properties. Understanding how to graph, transform, and solve equations and inequalities involving modulus is crucial for analyzing piecewise functions, absolute deviations, and error bounds in real-world applications.
In this guide: We'll master the definition and properties of modulus functions, learn to graph transformations of modulus functions, develop systematic techniques for solving modulus equations (often requiring case-by-case analysis), and tackle modulus inequalities using both algebraic and graphical methods essential for IB exams.
1. The Modulus Function
Definition
Modulus Function Definition
\(|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}\)
The modulus gives the non-negative value of \(x\)
Key Properties
Essential Properties:
- 1. Non-negativity: \(|x| \geq 0\) for all real \(x\)
- 2. Symmetry: \(|-x| = |x|\) (even function)
- 3. Triangle Inequality: \(|x + y| \leq |x| + |y|\)
- 4. Product: \(|xy| = |x| \cdot |y|\)
- 5. Quotient: \(\left|\frac{x}{y}\right| = \frac{|x|}{|y|}\) for \(y \neq 0\)
Graph of Modulus Function
For \(f(x) = |x|\):
- Domain: All real numbers, \(x \in \mathbb{R}\)
- Range: \(y \geq 0\) or \([0, \infty)\)
- Shape: V-shaped graph with vertex at origin \((0, 0)\)
- Symmetry: Symmetric about y-axis (even function)
- Gradient: -1 for \(x < 0\); +1 for \(x > 0\); undefined at \(x = 0\)
- Intercepts: x-intercept and y-intercept both at \((0, 0)\)
Transformations of Modulus
Two Important Cases:
Type 1: \(y = |f(x)|\)
Take the graph of \(f(x)\) and reflect any part below the x-axis upward
All negative y-values become positive
Example: \(y = |x - 2|\) reflects negative portions up
Type 2: \(y = f(|x|)\)
Keep the right side of graph (\(x \geq 0\)) and reflect it across the y-axis
Graph becomes symmetric about y-axis
Example: \(y = (|x|)^2 = x^2\) (already even, so unchanged)
⚠ Common Pitfalls:
- Forgetting negative case: \(|x|\) has two cases—don't forget \(x < 0\)!
- Sign confusion: When \(x < 0\), \(|x| = -x\) (which is positive!)
- Graph transformation: \(|f(x)|\) vs \(f(|x|)\) produce different graphs
- Range restriction: \(|f(x)|\) always has range \(\geq 0\)
2. Solving Modulus Equations
Basic Equation Types
Three Standard Forms:
Form 1: \(|x| = a\)
- If \(a > 0\): Two solutions, \(x = a\) or \(x = -a\)
- If \(a = 0\): One solution, \(x = 0\)
- If \(a < 0\): No solution (modulus cannot be negative)
Example: \(|x| = 5\) → \(x = 5\) or \(x = -5\)
Form 2: \(|f(x)| = a\)
Solve two equations: \(f(x) = a\) and \(f(x) = -a\)
Example: \(|x - 3| = 7\) → \(x - 3 = 7\) or \(x - 3 = -7\) → \(x = 10\) or \(x = -4\)
Form 3: \(|f(x)| = |g(x)|\)
Solve: \(f(x) = g(x)\) or \(f(x) = -g(x)\)
Example: \(|x + 1| = |2x - 3|\)
Case-by-Case Method
For Complex Modulus Equations:
- Identify critical points where expression inside modulus equals zero
- Split the domain into regions based on these points
- In each region, remove the modulus sign (positive or negative)
- Solve the resulting equation in each region
- Check solutions are valid for their region
- Combine all valid solutions
💡 Solving Tips:
- Always check your solutions by substituting back
- Remember \(|x| = a\) requires \(a \geq 0\)
- Square both sides only if both sides are already positive (to avoid extraneous solutions)
- Graphical method: find intersections of \(y = |f(x)|\) and \(y = a\)
Example 1: Solving Modulus Equations
Problem: Solve the following equations:
(a) \(|2x - 6| = 10\)
(b) \(|x - 1| = |2x + 3|\)
Solution:
(a) \(|2x - 6| = 10\)
This has form \(|f(x)| = a\) where \(a = 10 > 0\)
Solve two equations:
Case 1: \(2x - 6 = 10\)
\(2x = 16\)
\(x = 8\)
Case 2: \(2x - 6 = -10\)
\(2x = -4\)
\(x = -2\)
Check both solutions:
\(x = 8\): \(|2(8) - 6| = |10| = 10\) ✓
\(x = -2\): \(|2(-2) - 6| = |-10| = 10\) ✓
Solutions: \(x = 8\) or \(x = -2\)
(b) \(|x - 1| = |2x + 3|\)
This has form \(|f(x)| = |g(x)|\)
Solve two equations:
Case 1: \(x - 1 = 2x + 3\)
\(-1 - 3 = 2x - x\)
\(x = -4\)
Case 2: \(x - 1 = -(2x + 3)\)
\(x - 1 = -2x - 3\)
\(3x = -2\)
\(x = -\frac{2}{3}\)
Check both solutions:
\(x = -4\): \(|-5| = |2(-4) + 3| = |-5|\) → \(5 = 5\) ✓
\(x = -\frac{2}{3}\): \(|-\frac{5}{3}| = |\frac{-4}{3} + 3| = |\frac{5}{3}|\) → \(\frac{5}{3} = \frac{5}{3}\) ✓
Solutions: \(x = -4\) or \(x = -\frac{2}{3}\)
3. Solving Modulus Inequalities
Standard Inequality Forms
Key Inequality Rules
Type 1: \(|f(x)| < a\) (where \(a > 0\))
\(-a < f(x) < a\)
"The solution is between \(-a\) and \(a\)"
Example: \(|x| < 3\) → \(-3 < x < 3\)
Type 2: \(|f(x)| > a\) (where \(a > 0\))
\(f(x) < -a\) or \(f(x) > a\)
"The solution is outside the interval"
Example: \(|x| > 3\) → \(x < -3\) or \(x > 3\)
Special Cases:
- If \(a \leq 0\) and \(|f(x)| < a\): No solution
- If \(a < 0\) and \(|f(x)| > a\): All real numbers (since \(|f(x)| \geq 0\))
Solution Methods
Three Approaches:
Method 1: Algebraic (using inequalities above)
Apply the standard forms directly
Method 2: Case-by-case
Split into regions based on where expression = 0
Method 3: Graphical
Sketch \(y = |f(x)|\) and \(y = a\), find where inequality satisfied
⚠ Critical Mistakes:
- Wrong direction: \(|x| > 3\) is NOT \(-3 > x > 3\) (that's impossible!)
- Forgetting "or": \(|x| > 3\) means \(x < -3\) OR \(x > 3\) (two separate regions)
- Sign flipping: When multiplying/dividing by negative, flip inequality sign
- Not checking validity: Solutions must satisfy original inequality
Example 2: Solving Modulus Inequalities (IB-Style)
Problem: Solve the following inequalities:
(a) \(|x - 5| \leq 3\)
(b) \(|2x + 1| > 7\)
Solution:
(a) \(|x - 5| \leq 3\)
This has form \(|f(x)| \leq a\) where \(a = 3 > 0\)
Apply the rule: \(-a \leq f(x) \leq a\)
\(-3 \leq x - 5 \leq 3\)
Add 5 to all parts:
\(-3 + 5 \leq x \leq 3 + 5\)
\(2 \leq x \leq 8\)
Solution: \(x \in [2, 8]\) or \(2 \leq x \leq 8\)
Interpretation: \(x\) is within distance 3 from 5 on the number line
(b) \(|2x + 1| > 7\)
This has form \(|f(x)| > a\) where \(a = 7 > 0\)
Apply the rule: \(f(x) < -a\) or \(f(x) > a\)
\(2x + 1 < -7\) or \(2x + 1 > 7\)
Case 1: \(2x + 1 < -7\)
\(2x < -8\)
\(x < -4\)
Case 2: \(2x + 1 > 7\)
\(2x > 6\)
\(x > 3\)
Solution: \(x < -4\) or \(x > 3\)
Interval notation: \(x \in (-\infty, -4) \cup (3, \infty)\)
Interpretation: \(x\) is more than distance 7 from \(-\frac{1}{2}\) on the number line
📋 Modulus Quick Reference
| Type | Statement | Solution |
|---|---|---|
| Equation | \(|f(x)| = a\), \(a > 0\) | \(f(x) = a\) or \(f(x) = -a\) |
| Inequality < | \(|f(x)| < a\), \(a > 0\) | \(-a < f(x) < a\) |
| Inequality > | \(|f(x)| > a\), \(a > 0\) | \(f(x) < -a\) or \(f(x) > a\) |
| Double modulus | \(|f(x)| = |g(x)|\) | \(f(x) = g(x)\) or \(f(x) = -g(x)\) |
🎯 IB Exam Strategy
Common Question Types:
- "Solve \(|f(x)| = a\)": Two cases—positive and negative
- "Solve \(|f(x)| < a\)": Double inequality \(-a < f(x) < a\)
- "Solve \(|f(x)| > a\)": Two separate regions (OR statement)
- "Sketch \(y = |f(x)|\)": Reflect negative portions upward
- "Find where \(|f(x)| \leq a\)": Can use graphical interpretation
Key Reminders:
- Always check: does \(a \geq 0\)? (modulus cannot equal negative)
- For inequalities: "less than" gives AND, "greater than" gives OR
- Use GDC to verify graphically
- Check all solutions in original equation/inequality
🎉 Master Modulus Functions!
Modulus functions represent distance and magnitude—fundamental concepts in mathematics. Mastering their definition, graphical transformations, and systematic solution methods for equations and inequalities gives you powerful analytical tools essential for IB success and real-world applications!
Key Success Factors:
- ✓ Definition: \(|x| = x\) if \(x \geq 0\); \(|x| = -x\) if \(x < 0\)
- ✓ Equation \(|f(x)| = a\): solve \(f(x) = a\) and \(f(x) = -a\)
- ✓ Inequality \(|f(x)| < a\): gives \(-a < f(x) < a\) (AND)
- ✓ Inequality \(|f(x)| > a\): gives \(f(x) < -a\) or \(f(x) > a\) (OR)
- ✓ Graph \(y = |f(x)|\): reflect negative parts upward
- ✓ Always verify solutions work in original problem
Understand Distance • Split Into Cases • Check Solutions
Master modulus and excel in IB Mathematics! 🚀